LeetCode 690 Employee Importance 解题报告
You are given a data structure of employee information, which includes the employee's unique id, his importance value and his direct subordinates' id.
For example, employee 1 is the leader of employee 2, and employee 2 is the leader of employee 3. They have importance value 15, 10 and 5, respectively. Then employee 1 has a data structure like [1, 15, ], and employee 2 has [2, 10, ], and employee 3 has [3, 5, ]. Note that although employee 3 is also a subordinate of employee 1, the relationship is not direct.
Now given the employee information of a company, and an employee id, you need to return the total importance value of this employee and all his subordinates.
# Employee info
def __init__(self, id, importance, subordinates):
# It's the unique id of each node.
# unique id of this employee
self.id = id
# the importance value of this employee
self.importance = importance
# the id of direct subordinates
self.subordinates = subordinates
def getImportance(self, employees, id):
:type employees: Employee
:type id: int
ids = [employee.id for employee in employees]
leader = employees[ids.index(id)]
if leader.subordinates == :
im = 0
for id in leader.subordinates:
im += self.getImportance(employees, id)
return im + leader.importance